Question

The value of $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}$, where a, b, c, k are constants, depends only on
(a) a, b and c
(b) k
(c) a and b
(d) a and k

Answer 1

Hint: Integrate the given function and substitute the given values in the indefinite integration of the function. Solve them and check the parameters on which the integration depends. Use the fact that the value of $\int{a{{x}^{n}}dx}$ is $\dfrac{a{{x}^{n+1}}}{n+1}$. Another way to check the dependency of the integral on the parameters is by considering the fact that the integral of odd functions from ‘-a’ to ‘a’ is zero and those of even functions is positive.

Complete step-by-step answer:

We have to calculate the value of $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}$.
We will integrate the given function and substitute the given values and then simplify it to find the value of the integral.
We know that integration of sum of functions is equal to the sum of integration of each of the functions.
Thus, we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx.....\left( 1 \right)}$.
We also know that integration of function of the form $a{{x}^{n}}$ is given by $\int{a{{x}^{n}}dx}=\dfrac{a{{x}^{n+1}}}{n+1}$.
Thus, we have $\int{a{{x}^{5}}dx}=\dfrac{a{{x}^{6}}}{6}$.
Similarly, we have $\int{b{{x}^{3}}dx}=\dfrac{b{{x}^{4}}}{4}$.
We have $\int{cxdx}=\dfrac{c{{x}^{2}}}{2}$ and $\int{kdx=kx}$.
Substituting all the above integration in equation (1), we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\int_{-3}^{3}{a{{x}^{5}}dx}+\int_{-3}^{3}{b{{x}^{3}}dx}+\int_{-3}^{3}{cxdx}+\int_{-3}^{3}{kdx=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}}$.

We will now substitute the values $x=3$ and $x=-3$ in the integration of the function.

Thus, we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\left[ \dfrac{a{{x}^{6}}}{6}+\dfrac{b{{x}^{4}}}{4}+\dfrac{c{{x}^{2}}}{2}+kx \right]_{x=-3}^{x=3}=\dfrac{a{{\left( 3 \right)}^{6}}}{6}+\dfrac{b{{\left( 3 \right)}^{4}}}{4}+\dfrac{c{{\left( 3 \right)}^{2}}}{2}+k\left( 3 \right)-\left[ \dfrac{a{{\left( -3 \right)}^{6}}}{6}+\dfrac{b{{\left( -3 \right)}^{4}}}{4}+\dfrac{c{{\left( -3 \right)}^{2}}}{2}+k\left( -3 \right) \right]$.

Further simplifying the equation, we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( -3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( -3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( -3 \right)}^{2}} \right)+k\left( 3-\left( -3 \right) \right)$.

Thus, we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=\dfrac{a}{6}\left( {{3}^{6}}-{{\left( 3 \right)}^{6}} \right)+\dfrac{b}{4}\left( {{3}^{4}}-{{\left( 3 \right)}^{4}} \right)+\dfrac{c}{2}\left( {{3}^{2}}-{{\left( 3 \right)}^{2}} \right)+k\left( 3+3 \right)$.

So, we have $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}=0+k\left( 3+3 \right)=6k$.

Hence, the value of $\int_{-3}^{3}{\left( a{{x}^{5}}+b{{x}^{3}}+cx+k \right)dx}$ depends only on k, which is option (b).

Note: It’s not necessary to solve the integral completely and evaluate its value. We can also use the fact that the value of integration of odd functions from ‘-a’ to ‘a’ is zero and those of even functions is positive. Polynomials of odd degree are odd functions. Thus, the value of integration of $a{{x}^{5}}+b{{x}^{3}}+cx$ over the given range will be zero and non – zero for $\int_{-3}^{3}{kdx}$. Hence, the total value of integration will be dependent on k.